Hydrodynamics : Calculate force exerted by fluid on Tube

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SUMMARY

The discussion focuses on calculating the net force exerted by a fluid on a tube bent into a circular arc, specifically three-quarters of a circle. The user derived the force using the formula F = (3πAρv²)/2, but the textbook answer is √2 ρAv². The discrepancy arises from the user's approach of summing outward radial forces without considering vector directions. The correct method involves analyzing the change in momentum based on the flow direction at the tube's entry and exit points.

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Homework Statement



A tube of uniform cross-section A is bent to form a circular arc of radius R, forming three quarters of a circle. A liquid of density \rho is forced through the tube with a linear speed v. Find the net force exerted by the liquid on the tube.

2. The attempt at a solution

The centripetal force acting on a small element of dm mass will be

dF=\frac{v^2}{R}dm=\frac{\rho v^2}{R}dV=\frac{\rho A v^2}{R}dl.

Integrating with proper limits, I got

F=\frac{\rho A v^2}{R}\int_{0}^{3\pi R/2}dl=\frac{3\pi A \rho v^2}{2}

However the answer provided in my book is \sqrt{2} \rho Av^2 .What am I doing wrong. Can someone guide me?
 
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It says 'net force', but you added all the outward radial forces by magnitude, not as vectors. In fact, the shape doesn't matter. All you need to know is the direction of the flow at entry and exit. What is the change in momentum?
 

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